• + 3 comments

    O(300^2 + n * m^2)

    long mod = 1000000007;
    
    void combinatorial(vector<vector<long>>& binomial) {
        binomial[1][0] = 1;
        binomial[1][1] = 1;
        for (int n = 2; n <= 300; n++) {
            binomial[n][0] = 1;
            for (int k = 1; k <= n; k++) binomial[n][k] = (binomial[n-1][k-1] + binomial[n-1][k]) % mod;
        }
    }
    
    int gridWalking(int n, int m, const vector<int>& x, const vector<int>& D, const vector<vector<long>>& binomial) {
        vector<vector<long>> oneDimension(n+1, vector<long>(m+1)), multiDimension(n+1, vector<long>(m+1)), endPoint = {{0}};
        for (int i = 1; i <= n; i++) {
            oneDimension[i][0] = 1;
            endPoint.emplace_back(vector<long>(D[i-1] + 2));
            endPoint.back()[x[i-1]] = 1;
            for (int K = 1; K <= m; K++) {
                long total = 0, temp1 = 0, temp2;
                for (int point = 1; point <= D[i-1]; point++) {
                    temp2 = endPoint[i][point];
                    endPoint[i][point] = (temp1 + endPoint[i][point + 1]) % mod;
                    total = (total + endPoint[i][point]) % mod;
                    temp1 = temp2;
                }
                oneDimension[i][K] = total;
            }
        }
        multiDimension[1] = oneDimension[1];
        for (int i = 2; i <= n; i++) {
            multiDimension[i][0] = 1;
            for (int K = 1; K <= m; K++) {
                long temp = 0;
                for (int l = 0; l <= K; l++) temp = (temp + multiDimension[i-1][l] * (oneDimension[i][K-l] * binomial[K][l] % mod)) % mod;
                multiDimension[i][K] = temp;
            }
        }
        return multiDimension[n][m];
    }
    
    int main()
    {
        int t, n, m, k;
        vector<vector<long>> binomial(301, vector<long>(301));
        combinatorial(binomial);
        cin >> t;
        for (int i = 1; i <= t; i++) {
            vector<int> x, D;
            cin >> n >> m;
            for (int j = 1; j <= n; j++) {
                cin >> k;
                x.push_back(k);
            }
            for (int j = 1; j <= n; j++) {
                cin >> k;
                D.push_back(k);
            }
            cout << gridWalking(n, m, x, D, binomial) << '\n';
        }
    }
    
    • + 1 comment

      why binomial[n][k] = (binomial[n-1][k-1] + binomial[n-1][k]) % mod;?

      • + 0 comments

        lol i dont remember how i did this question. this is wat i wrote as explanation, it explains the whole thing, i cant be bothered to read it to explain ur specific question

        //A random walk in n dimensions can be split into random walk in each of the dimensions and analysed separately

        //First, for each dimension i and 1 <= K <= m, oneDimension[i][K] calculates the number of random walks of length K //within the interval [1, D[i]]: //(1) For 1 <= point <= D[i], endPoint[i][K][point] stores the the number of length K random walks within [1, D[i]] that // ends at point //(2) endPoint[i][K+1][point] = endPoint[i][K][point-1] + endPoint[i][K][point+1] //(3) oneDimension[i][K] = sum of all the endPoint[i][K][point] for every 1 <= point <= D[i]

        //For n dimensions, let multiDimension[n][K] be the number of random walks within the corresponding n-dimensional cube defined by //the D[i]'s //The dynamic programming part is: any such walk can be split into a walk A in dimension n, and a walk B in the remaining n-1 dimensions.

        //Let binomial[n][k] be the binomial coefficient //Given a walk A of length l in dimension n, and a walk B of length K-l in the remaining n-1 dimensions, //they can be combined to become a walk H of length K in n dimensions. //There's binomial[K][l] many ways to insert the walk A into H, hence the number of length K walk in n dimensions that //can be decomposed into a length l walk in dimension n, and a length K-l walk in the remaining n-1 dimensions, is: // oneDimension[n][l] * multiDimension[n-1][K-l] * binomial[K][l] //=> multiDimension[n+1][K] = sum of all the oneDimension[n][l] * multiDimension[n-1][K-l] * binomial[K][l] for every 0 <= l <= K

        //The time of this algorithm is O(300^2 + n * m^2)

    • + 0 comments

      why x[i-1] not x[i]?

    • + 0 comments

      very beatiful solution